Show that $\frac{d^n}{dx^n}(\frac{\log x}{x})=(-1)^n\frac{n!}{x^{n+1}}(\log x-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n})$

Question

Show that $\frac{d^n}{dx^n}\left(\frac{\log x}{x}\right)=(-1)^n\frac{n!}{x^{n+1}}\left(\log x-1-\frac{1}{2}-\frac{1}{3}-\ldots -\frac{1}{n}\right)$

I am not allowed to use induction. I do not know how to approach this. Any help would be appreciated.

Answer 1

We know $$\frac1{x+h}=\frac1x\frac1{1+h/x}=\sum^\infty_{n=0}\frac{(-1)^n}{x^{n+1}}\,h^n=\sum^\infty_{n=0}a_n(x)\,h^n,$$ and by integration $$\log(x+h)=\log x+\int^h_0\frac1{x+h}\,dh=\log x+\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n\,x^n}\,h^n=\sum^\infty_{n=0}b_n(x)\,h^n.$$ Then (Cauchy product of power series) $$\frac{\log(x+h)}{x+h}=\sum^\infty_{n=0}c_n(x)\,h^n,$$ where $$c_n(x)=\sum^n_{k=0}a_k(x)\,b_{n-k}(x)=\sum^{n-1}_{k=0}\frac{(-1)^k}{x^{k+1}}\frac{(-1)^{n-k-1}}{(n-k)\,x^{n-k}}+\frac{(-1)^n}{x^{n+1}}\,\log x,$$ meaning $$c_n(x)=\frac{(-1)^n}{x^{n+1}}\,\left(\log x-1-\frac12-\ldots-\frac1n\right).$$ But the coefficient at $h^n$ of this Taylor series is $$\frac1{n!}\frac{d^n}{dx^n}\frac{\log x}x.$$

Answer 2

Another proof

Let $f(x) = \frac{\log(x)}{x}$, then for $x, y > 0$ $$f(x y) = \frac{\log(x)+ \log(y)}{x y} = \frac{f(x)}{y} + \frac{f(y)}{x}$$ Derivating $n$ times with respect to $y$ yields $$x^n f^{(n)}(x y) = f(x) \left(\frac{1}{y}\right)^{(n)} + \frac{f^{(n)}(y)}{x}$$ Now replacing $y$ by $1$ yields $$x^n f^{(n)}(x) = f(x) (-1)^n n! + \frac{f^{(n)}(1)}{x}$$ hence $$f^{(n)}(x) = (-1)^n n! \frac{\log(x)}{x^{n+1}} + \frac{f^{(n)}(1)}{x^{n+1}}$$

The value of $f^{(n)}(1)$ is missing, it can be obtained by the Taylor series of $$f(1+h) = \frac{\log(1+h)}{1+h}$$ used by Professor Vector, but there is another way: rewrite the above formula as $$f^{(n)}(x) = (-1)^{n}n! \frac{\log(x) - s_n}{x^{n+1}}$$ We have $s_0=0$ and we claim that $s_{n+1} - s_n = \frac{1}{n+1}$ from which it follows that $$s_n = s_n - s_0 = \sum_{k=0}^{n-1}(s_{k+1}-s_k) = \sum_{k=0}^{n-1}\frac{1}{k+1} = 1 + \frac{1}{2} +\cdots+\frac{1}{n}$$ To prove our claim, we compute $$(-1)^{n+1}(n+1)! \frac{\log(x) - s_{n+1}}{x^{n+2}} = f^{n+1}(x) = (f^{(n)}(x))^\prime =(-1)^{n}n! \frac{1 -(n+1)(\log(x) - s_n)}{x^{n+2}}$$ It follows that $s_{n+1} = \frac{1}{n+1} + s_n$, QED

Answer 3

Approach 1: When we calculate the first few derivatives, we have:

$$y'=-\frac{1}{x^2}\ln{x}+\frac{1}{x^2}=-\frac{1}{x^2}(\ln{x}-1).$$ $$y''=\frac{2}{x^3}\left(\ln{x}-1-\frac12 \right).$$ $$y^{(3)} = -\frac{3!}{x^4}\left(\ln{x}-1-\frac12 -\frac13\right).$$

so we can recognize the desired pattern.

Approach 2: We use the generalised Leibniz rule:

$$(fg)^{n} = \sum_{k= 0}^n \binom{n}{k} f^{(n-k)}(x) g^{(k)}(x)$$

with $f(x) = 1/x, g(x) = \log x$

to obtain:

$$\left(\frac{\log x}{x}\right)^{(n)}= \sum_{k= 0}^n \binom{n}{k} \log^{(k)}(x) \left(\frac{1}{x}\right)^{(n-k)}(x)$$ $$= \sum_{k= 0}^n \binom{n}{k} \log^{(k)}(x) \frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$ $$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \binom{n}{k} \left(\frac{1}{x}\right)^{(k-1)} \frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$

$$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \binom{n}{k} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$

$$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k!(n-k)!} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$

$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k(k-1)!(n-k)!} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$

$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k} \frac{(-1)^{k-1}}{x^{k}}\frac{(-1)^{n-k}}{x^{n-k+1}}$$

$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \frac{n!}{k} \frac{(-1)^{n-1}}{x^{n+1}}$$

$$ = \log(x) \frac{(-1)^{n}n!}{x^{n+1}} - \frac{n!(-1)^n}{x^{n+1}} \sum_{k= 1}^n \frac{1}{k}$$

$$= \frac{n!(-1)^n}{x^{n+1}} \left(\log(x) - 1 - \frac{1}{2} - \dots - \frac{1}{n}\right)$$

Approach 3

Induction: For $n=0$, the formula holds.

Assume it holds for $n-1$, and then show it holds for $n$, by taking the derivative and rearranging terms.

Other/similar approaches: n-th derivative of $\frac{\ln x}{x}$.